1. **Problem Statement:** Find all possible pairs $(a,b)$ such that the number $48a23b$ is a multiple of 18. 2. **Key Concept:** A number is a multiple of 18 if and only if it is divisible by both 2 and 9. 3. **Divisibility by 2:** The last digit $b$ must be even. So, $b \in \{0,2,4,6,8\}$. 4. **Divisibility by 9:** The sum of the digits must be divisible by 9. Sum of digits = $4 + 8 + a + 2 + 3 + b = 17 + a + b$. 5. **Check each possible $b$ and find $a$ such that $17 + a + b$ is divisible by 9:** - For $b=0$: $17 + a + 0 = 17 + a$ divisible by 9. Closest multiples of 9 near 17 are 18 and 27. $a=1$ makes sum 18. - For $b=2$: $17 + a + 2 = 19 + a$ divisible by 9. Closest multiples: 18, 27. $a=8$ makes sum 27. - For $b=4$: $17 + a + 4 = 21 + a$ divisible by 9. Closest multiples: 18, 27. $a=6$ makes sum 27. - For $b=6$: $17 + a + 6 = 23 + a$ divisible by 9. Closest multiples: 18, 27. $a=4$ makes sum 27. - For $b=8$: $17 + a + 8 = 25 + a$ divisible by 9. Closest multiples: 18, 27. $a=2$ makes sum 27. 6. **Final pairs:** $(a,b) = (1,0), (8,2), (6,4), (4,6), (2,8)$. **Answer:** The possible pairs $(a,b)$ are $(1,0), (8,2), (6,4), (4,6), (2,8)$.